Carcass wrote:
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability
that neither of the lightbulbs selected will be defective?
Give your answer as a fraction.
Practice Questions
Question: 24
Page: 344
Difficulty: medium
P(neither lightbulb is defective) = P(BOTH lightbulbs work)
= P(1st lightbulb works
AND 2nd lightbulb works)
= P(1st lightbulb works)
x P(2nd lightbulb works)
= 18/20
x 17/19
= 9/10
x 17/19
[since 18/20 = 9/10]= 153/190
Answer: 153/190
ASIDE: P(1st lightbulb works) = 18/20, because there are 20 lightbulbs and 18 of them work.
P(2nd lightbulb works) = 17/19 because,
once we have selected a working bulb for the 1st selection, there are 19 lightbulbs remaining and 17 of them work.
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep